242 ANALYTICAL MECHANICS
where M is the total mass and x, y, and z are the coordinates of the center of mass of the system. Combining the last three equations in a single vector equation we obtain
2mv = Mv; (VII)
which states that the resultant momentum of a system equals the product of the total mass of the system by the velocity of its center of mass.
196. Motion of the Center of Mass of a System. — Combining equations (V) and (VII) we get
F = Mv, (VIII)
which states that the resultant external force acting upon a system equals the product of the total mass of the system by the acceleration of its center of mass. But equation (VIII) is the force equation for a particle of* mass M, which is acted upon by a force F. Therefore the center of mass of a system moves as if the entire, mass of the system were concentrated at that point and all the forces acting upon the system were applied to the resulting particle.
PROBLEM.
Show that when the component, along any direction, of the resultant force acting upon a system vanishes the corresponding component of the velocity of the center of mass remains constant, that is,
x — const., when X = 0.
ILLUSTRATIVE PROBLEM.
A bullet penetrates a fixed plate to a depth d. How far would it penetrate if the plate were free to move in the direction of motion of the bullet?
Let F be the mean resisting force which the plate offers to the motion ^f the bullet. When the plate is fixed all the energy of the bullet is ex-_pended in doing work against this force. Therefore we have
^ Fd = i mi;2, (1)
where m is the mass and v the velocity of the bullet. When the target is free to move part of the energy of the bullet is expended in giving the